Answer to Question #95824 in Statistics and Probability for Sam

Question

Many manufacturers have quality control programs that include inspection of incoming materials for defects. Suppose that a computer manufacturer receives computer boards in lots of five. Two boards are randomly selected without replacement from each lot for inspection. We can represent possible outcomes of the selection process by pairs. For example, the pair (1,2) represents the selection of boards 1 and 2 for inspection regardless of the order of selection. Define random variable X to be the number of defective boards observed among the two inspected.
(a) By listing the ten different possible outcomes, find the probability distribution of X assuming that boards 1 and 2 are the only defective boards in a lot of five.
(b) Calculate µX and σX.

Answer 1

There are total 5 batches and two boards are selected from each batch for inspection.

Let the boards are numbered from 1 to 5.

If the selected boards are 1 and 2, then it is represented in pair as (1, 2).

If the selected boards are 1 and 3, then it is represented in pair as (1, 3).

Similarly, other pairs can be obtained.

a) Let X be the number of defective boards observed among the two inspected.

If the boards 1 and 2 are the only defective boards in a lot of five, then

$(1, 2), x=2; \ (1,3),x=1;\ (1,4), x=1;\ (1,5),x=1;$

 $(2, 3), x=1; \ (2,4),x=1;\ (2,5), x=1;$

$(3,4),x=0;\ (3,5),x=0;\ (4,5),x=0.$

$$P(X=0)={3 \over 10}=0.3$$ $$P(X=1)={6 \over 10}=0.6$$ $$P(X=2)={1 \over 10}=0.1$$

b)

$$\begin{matrix} x & 0 & 1 & 2 \\ p(x) & 0.3 & 0.6 & 0.1 \end{matrix}$$

$$\mu_X=\displaystyle\sum_{i=1}^nx_ip(x_i)$$

$$\mu_X=0\cdot0.3+1\cdot0.6+2\cdot 0.1=0.8$$

$$Var(X)=\sigma_X^2=\displaystyle\sum_{i=1}^n(x_i-\mu_X)^2p(x_i)$$ $$\sigma_X^2=(0-0.8)^2\cdot0.3+(1-0.8)^2\cdot0.6+(2-0.8)^2\cdot0.1=0.36$$ $$\sigma_X=\sqrt{\sigma_X^2}=\sqrt{0.36}=0.6$$

$\mu_X=0.8, \ \sigma_X=0.6$